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Question

The electric field in a region is given by E=25Ei^+35Ej^ with E=4.0×103N/C.The flux of this field through a rectangular surface are 0.4m2parallel to Y–Z plane is________Nm2C-1.


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Solution

Step 1: Given data

The electric field in the region, E=25Ei^+35Ej^

Intensity E=4.0×103N/C

Area=0.4m2

Step 2: Formula Used

Gauss’ law can be used to compute the flux. The following expression represents the law,

ϕ=EdA----1

Step 3: Compute the flux

Consider the following figure:

The plane is parallel to the yz-plane. Hence only 25Ei passes perpendicular to the plane whereas35Ej goes parallel.

According to the figure, substitute the known values in the equation 1,

ϕ=25E×0.4ϕ=25×4.0×103×0.4ϕ=640Nm2c-1

Hence, the flux of this field through a rectangular surface will be 640Nm2c-1.


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